Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What is the pH of the solution created by combining 11.80 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)?
with 8.00 mL of the 0.10 M HC2H3O2(aq)?

Complete the table below: Note: Make simplifying assumptions, do not use the quadratic formula.
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?

..........HAc + NaOH ==> NaAc + H2O
initial..0.800..1.180.....0......0
change..-0.800..-0.800..0.800..0.800
equil....0......0.380...0.800..0.800
Ignoring the OH^- that might be added due to the hydrolysis of the NaAc salt (which is negligible) pH is determined by the mmoles NaOH/mL.

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